Question

some states now allow online gambling. as a marketing manager for a casino, you need to determine the percentage of adults in those states who gamble online how many adults must you survey in order to be 90% confident that your estimate is in error by no more than five percentage points? complete parts (a) and (b) below

a. assume that nothing is known about the percentage of adults who gamble online.

( n=square )

(round up to the nearest integer)

b. assume that ( 19 % ) of all adults gamble online.

( n=square )

(round up to the nearest integer)

Explanation:

Step1: Find the z - value

For a 90% confidence level, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2=0.05\). Using the standard normal distribution table, \(z_{\alpha/2}=z_{0.05}\approx1.645\)

Step2: Calculate the sample size when no prior estimate is known (part a)

When no prior estimate \(p\) is known, we use \(p = 0.5\) (this maximizes the value of \(p(1 - p)\)). The margin of error \(E = 0.05\).
The formula for the sample size \(n\) in estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}\times p(1 - p)}{E^{2}}\)
Substitute \(z_{\alpha/2}=1.645\), \(p = 0.5\), \(1-p=0.5\) and \(E = 0.05\) into the formula:

$$
LATEXBLOCK0
$$

Round up to \(n = 271\)

Step3: Calculate the sample size when \(p=0.19\) (part b)

Given \(p = 0.19\), then \(1 - p=1 - 0.19 = 0.81\), \(z_{\alpha/2}=1.645\) and \(E = 0.05\)

$$
LATEXBLOCK1
$$

Round up to \(n = 167\)

Answer:

a. \(271\)
b. \(167\)